Beginning at 8AM customers arrive at the bank at regular exponential intervals. Vivian arrives at the bank at 11AM and, being a VIP customer, is promoted to the head of the line. How long will she have to wait in line (i.e. what is the distribution of her waiting time)?
Formally, let $\left(\Omega, \mathcal{F}, P\right)$ be a probability space, and let $\left(X_1, X_2,\dots\right)$ be an infinite sequence of i.i.d random variables defined in this space with $X_1\sim\mathrm{Exp}\left(\lambda\right)$, for some $\lambda\in\left(0,\infty\right)$. Denote by $\left(S_1, S_2, \dots\right)$ the sequence of partial sums (i.e. for any $n\in\mathbb{N}_1$, $S_n = \sum_{k=1}^nX_k$), and set $S_0 := 0\mathbb{1}_\Omega$. Letting $c\in\left(0,\infty\right)$, define $\tau:\Omega\rightarrow\mathbb{N}_0$ to be the stopping time $\tau\left(\omega\right) := \arg\max_{n\in\mathbb{N}_0}S_n\left(\omega\right)< c$. Finally, define $Y:= S_{\tau+1} - c$. What is $Y$'s distribution?
My guess is $Y\sim\mathrm{Exp}\left(\lambda\right)$, due to the exponential distribution's memorilessness, but i can't figure out how to prove it formally.
The "real life" situation in the first part of the post is not adequate since there could be zero customers in line when the VIP customer arrives. The mathematical question in the second part of the post is classical renewal theory, the general result being as follows:
When $X$ is exponential $\lambda$, $\hat X$ is gamma $(2,\lambda)$ and $U\hat X$ is exponential $\lambda$, just like $X$. In general, the distribution of $U\hat X$ has density $f:x\mapsto P[X\geqslant x]/E[X]$ hence exponential distributions are the only ones such that $U\hat X$ is distributed like $X$.